When 5.12g of NaOH were dissolved in 51.55g water in a calorimeter at 24.5C, the

Question

When 5.12g of NaOH were dissolved in 51.55g water in a calorimeter at 24.5C, the temperature of the solution went up to 49.8C a) calculate qh20 b) find delta H for the reaction as it occurred in the calorimeter c) find delta H for the solution of 1g NaOH in water d) find delta H for the solution of 1 mole NaOH in water e) Given the following heats of formation, delta Hf, in kJ per mole, as obtained from a table of delta Hf data, calculate delta H for the reaction NaOH(s)---->Na+(aq)+OH-(aq). Compare your answer with the result you obtained for part d. NaOH(s), -425.6; Na+(aq), -240.1; OH-, -230.0

Explanation / Answer

a)Yes, it releases heat, the temperature of the solution goes from about 24ºC to about 49ºC. b)qH2O= sensible heat of H2O= heat that causes H2O's temperature to change= q=m*c*?t q=51.55*1*(49.8 - 24.5) q=1,304 cal c)The change H?? Didn't understand. Anyway, that's the reaction: NaOH(s) + H2O(l) => Na+ + OH- + H20 + HEAT . Source(s): q=m*c*?t is the formula for calculating sensible heats q means heat and its unit is "cal" (calories) m is mass and it's unit g (gram) c is sensible heat and its unit is cal/ g * °C ? means variation and t means temperature,

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